Question: $\sum\limits_{n=1}^{\infty } \dfrac{(n+2)(x+5)^n}{n\cdot6^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-8 \le x<4$ (Choice B) B $-11 \le x<1$ (Choice C) C $-8<x<4$ (Choice D) D $-11<x<1$
Solution: We use the ratio test. For $x\neq -5$, let $a_n= \dfrac{(n+2)(x+5)^n}{n\cdot6^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+5}{6}\right| $ The series converges when $\left| \dfrac{x+5}{6}\right| <1$, which is when $-11<x<1$. Now let's check the endpoints, $x=-11$ and $x=1$. Letting $x=-11$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(n+2)(-11+5)^n}{n\cdot6^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(n+2)(-6)^n}{n\cdot6^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot(n+2)\cdot 6^n}{n\cdot6^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot(n+2)}{n} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. Letting $x=1$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(n+2)(1+5)^n}{n\cdot6^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(n+2)\cdot6^n}{n\cdot6^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{(n+2)}{n} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. In conclusion, the interval of convergence is $-11<x<1$.